Find the solution of the system
x' = 3x + 4y; y' = -2x-3y
satisfying x(0) = 2 and y(0) = -1
I turned the pair of equations into a single first order vector differential equation of the form
v' = Av
where v is a column vector with entries x and y and A is a 2x2 matrix.
A= 3 4
-2 -3
=>det(A-Ir),where r is a constant and I is the identety matrix....=(3-r)(-3-r)+8=r^2-1=>
r1=-1 and r2=1 these are the eigenvectors.Was my approach ok?
Finding egenvectors:
(A-Ir)w_j=|3-1 4 | w1 = 0
|-2 -3+1 | w2 0
this result the system 2w_1+4w_2=0 and -2w_1-2w_2=0 where both w's =0
so the general solution is u=a_1 e^t *|0| +a_2e^-t|0|
|0| |0|
I know i did a mistake here ,was r1=i and r2=-i? ..adn yes I have to use eigenvectors and eigenvalues

February 6th 2011, 07:44 AM

TheEmptySet

Quote:

Originally Posted by PatrickM

Find the solution of the system
x' = 3x + 4y; y' = -2x-3y
satisfying x(0) = 2 and y(0) = -1
I turned the pair of equations into a single first order vector differential equation of the form
v' = Av
where v is a column vector with entries x and y and A is a 2x2 matrix.
A= 3 4
-2 -3
=>det(A-Ir),where r is a constant and I is the identety matrix....=(3-r)(-3-r)+8=r^2-1=>
r1=-1 and r2=1 these are the eigenvectors.Was my approach ok?
Finding egenvectors:
(A-Ir)w_j=|3-1 4 | w1 = 0
|-2 -3+1 | w2 0
this result the system 2w_1+4w_2=0 and -2w_1-2w_2=0 where both w's =0
so the general solution is u=a_1 e^t *|0| +a_2e^-t|0|
|0| |0|
I know i did a mistake here ,was r1=i and r2=-i? ..adn yes I have to use eigenvectors and eigenvalues